Problem: The polynomial $p(x)=x^3+3x^2-4$ has a known factor of $(x-1)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Solution: We know $(x-1)$ is a factor of $p(x)$. This means that $p(x)=(x-1)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x-1)$ Notice that $p(x)$ is missing a $1^{\text{st}}$ degree term. Let's add it as $0x$. $\begin{array}{r} x^2+4x+4 \\ x-1|\overline{x^3+3x^2+0x-4} \\ \mathllap{-(}\underline{x^3-\phantom{4}x^2\phantom{+0x-4}\rlap)} \\ 4x^2+0x-4 \\ \mathllap{-(}\underline{4x^2-4x\phantom{-4}\rlap)} \\ 4x-4 \\ \mathllap{-(}\underline{4x-4\rlap)} \\ 0 \end{array}$ We find that $q(x)=x^2+4x+4$. Factoring $q(x)$ We can factor $x^2+{4}x+{4}$ as $(x+m)(x+n)$ where $m+n={4}$ and $m\cdot n={4}$. Such numbers are $2$ and $2$, so the factored expression is $(x+2)(x+2)$. Putting it all together $\begin{aligned} p(x)&=x^3+3x^2-4 \\\\ &=(x-1)(x^2+4x+4) \\\\ &=(x-1)(x+2)(x+2) \end{aligned}$ Notice that we can also write $p(x)$ as $(x-1)(x+2)^2$. This is still a product of linear factors!